Problem: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $10.5$ years; the standard deviation is $1.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living longer than $13.7$ years.
Explanation: $10.5$ $8.9$ $12.1$ $7.3$ $13.7$ $5.7$ $15.3$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $10.5$ years. We know the standard deviation is $1.6$ years, so one standard deviation below the mean is $8.9$ years and one standard deviation above the mean is $12.1$ years. Two standard deviations below the mean is $7.3$ years and two standard deviations above the mean is $13.7$ years. Three standard deviations below the mean is $5.7$ years and three standard deviations above the mean is $15.3$ years. We are interested in the probability of a meerkat living longer than $13.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the meerkats will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $7.3$ years and the other half $({2.5\%})$ will live longer than $13.7$ years. The probability of a particular meerkat living longer than $13.7$ years is ${2.5\%}$.